uniform convergence $\sum _{n=0} ^{\infty} \frac{\log (1+nx)}{nx^n}$ on $x \in (1,\infty)$

Question

How could I prove that $\sum _{n=0} ^{\infty} \frac{\log (1+nx)}{nx^n}$ is not uniformly convergent on an interval $I=(1,+\infty)$?
So far I have been thinking of proving that $f_n(x)=\frac{\log (1+nx)}{nx^n}$ does not uniformly converge on the interval, which leads me to finding $\lim _{n\to \infty}\sup_{x>1} \frac{\log (1+nx)}{nx^n}$. But there could be a nicer way.

Answer 1

It's easy to see that each summand in the series is bounded on $(1,\infty).$ If the series were uniformly convergent on $(1,\infty),$ then the series would sum to a bounded function there.

To show this fails, note that

$$\ln(1+nx) > \ln (nx) = \ln n + \ln x > \ln n.$$

Thus our series is bounded below by

$$\tag 1\sum_{n=1}^{\infty}\frac{\ln n}{nx^n}.$$

We're done if we show that as $x\to 1^+,$ $(1)\to \infty.$ That is easy to do, since $\sum (\ln n)/n=\infty.$

Answer 2

For large enough $N$ consider the inequalities \begin{align} x^{-N-1}\,\Phi\left(\frac{1}{x},1,N+1\right) &= x^{-N-1}\sum_{n=0}^\infty \frac{x^{-n}}{N+1+n}=\sum_{n=N+1}^\infty \frac{x^{-n}}{n} \\ &\leq \sum_{n=N+1}^\infty \frac{\log(1+nx)}{nx^n} \leq \sum_{n=N+1}^\infty x^{-n+1} = \frac{x^{-N}}{1-\frac{1}{x}} \end{align} where $\Phi$ is the Lerch-Zeta function which has the expansion $$ \Phi(z,s,a)=\frac1{2\,a^s}+\frac{\log^{s-1}\dfrac1z}{z^a}\,\Gamma(1-s,a\log\dfrac1z)+\frac2{a^{s-1}}\int\limits_0^\infty \frac{\sin(s\arctan t-a\,t\log z)}{(1+t^2)^{s/2}\cdot (\mathrm{e}^{2\,\pi\,a\,t}-1)}\,\mathrm dt $$ i.e. for $z=\frac{1}{x}$, $s=1$ and $a=N+1$ $$ \Phi\left(\frac{1}{x},1,N+1\right) = \frac{1}{2(N+1)} - x^{N+1} \, {\rm Li}\left(x^{-N-1}\right) + 2\int_0^\infty \frac{\sin\left(\arctan(t) - t\log\left(x^{-N-1}\right)\right)}{\left(e^{2\pi t(N+1)}-1\right)\sqrt{1+t^2}} \, {\rm d}t \, . $$ About $x=1$ the logarithmic integral is $$ {\rm Li}\left(x^{-N-1}\right) = {\rm Ei}\left(\log\left(x^{-N-1}\right)\right) = - {\rm Ei}_1\left(-\log\left(x^{-N-1}\right)\right) = \gamma + \log(N+1) + \log(x-1) + {\cal O}\left(x-1\right) $$ in fact for $x>1$ we have $$ \gamma + \log(N+1) + \log(x-1) \geq {\rm Li}\left(x^{-N-1}\right) \, . $$ For $x \in (1,2)$ the last integral is bounded by some constant $C$ and choose the minimal value on $(1,2)$.

Take any $\epsilon >0$ and $x_0 \in (1,2)$ and set $$ N=\left\lceil \frac{\log\left(\epsilon\left(1-\frac{1}{x_0}\right)\right)}{-\log \left( x_0 \right)} \right\rceil $$ for then we have $$ \sum_{n=m}^\infty \frac{\log(1+nx_0)}{nx_0^n} < \epsilon \qquad \forall m\geq N+1 \, . $$

Now take $x\in(1,2)$ sufficiently smaller $x_0$ when we have $$ x^{-N-1} \left(\frac{1}{2(N+1)} + C \right) - \gamma - \log(N+1) - \log(x-1) \leq x^{-N-1}\,\Phi\left(\frac{1}{x},1,N+1\right) $$ for the LHS. Therefore $$ \sum_{n=N+1}^\infty \frac{\log(1+nx')}{nx'^n} > \epsilon \qquad \forall x'\leq x \, . $$